Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int g[110][110], row[110], col[110];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) > 0) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        scanf("%d", &g[i][j]);
      }
    }
    int r0, best = (1 << 30);
    for (row[0] = 0; true; row[0]++) {
      bool test = true;
      int ans = row[0];
      for (int j = 0; j < m; j++) {
        if (g[0][j] < row[0]) {
          test = false;
          break;
        }
        col[j] = g[0][j] - row[0];
        ans += col[j];
      }
      if (!test) break;
      for (int i = 1; i < n and test; i++) {
        if (g[i][0] < col[0]) {
          test = false;
          break;
        }
        row[i] = g[i][0] - col[0];
        ans += row[i];
        for (int j = 1; j < m; j++) {
          if (row[i] + col[j] != g[i][j]) {
            test = false;
            break;
          }
        }
      }
      if (test and ans < best) {
        best = ans;
        r0 = row[0];
      }
    }
    if (best == (1 << 30)) {
      printf("-1\n");
      continue;
    }
    printf("%d\n", best);
    row[0] = r0;
    for (int j = 0; j < m; j++) {
      col[j] = g[0][j] - row[0];
    }
    for (int i = 1; i < n; i++) {
      row[i] = g[i][0] - col[0];
    }
    for (int i = 0; i < n; i++) {
      while (row[i]--) printf("row %d\n", i + 1);
    }
    for (int j = 0; j < m; j++) {
      while (col[j]--) printf("col %d\n", j + 1);
    }
  }
}